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Specific Heat Capacity

The energy re...

Question

The energy released in the reaction $_{1} H^{2} +_{1} H^{3}$ $\to_{2} H e^{4} +_{0} n^{1}$ is $- ------- -$
mass of $_{0} n^{1} = 1.00867 a m u$
mass of $_{1} H^{2} = 2.01410 a m u$
mass of $_{1} H^{3} = 3.0160 a m u$
mass of $_{2} H e^{4} = 4.0026 a m u$

0.01833 M e V

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17.53 M e V

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17.53 j o u l e

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0.01833 j o u l e

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Solution

The correct option is B $17.53 M e V$
$Δ m = [m_{_{1} H^{2}} + m_{_{1} H^{3}} - m_{_{2} {H e}^{4}} - m_{_{0} n^{1}}]$
$= [2.01410 + 3.0160 - 4.0026 - 1.00876] μ$
$= 0.01883 μ$
$Q = 0.01883 \times 931.478 M e V$
$= 17.53 M e V$

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Similar questions

_{1} H^{2} +

_{1} H^{3}

⟶

_{0} n^{1} +

_{2} H e^{4} .

1_{H}^{1}

20 MeV

_{1} H^{2} +_{1} H^{3} \to_{0} n^{1} +_{2} {He}^{4}

_{1} H^{2}

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10^{40}

_{1} H^{2} +_{1} H^{2} \to_{1} H^{3} +_{1} H^{1}

_{1} H^{2} +_{1} H^{3} \to_{2} H e^{4} +_{0} n^{1}

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_{1} H^{2} +_{0} n^{1} \to_{2} H e^{4} +_{0} n^{1}

m_{1} H^{1} = 1.007825 a m u

m_{1} H^{2} = 2.014102 a m u

m_{1} H^{3} = 3.016049 a m u

_{1} H^{2} +_{1} H^{2} \to_{1} H^{3} +_{1} H^{1}

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